Answer:
The momentum is 1.94 kg m/s.
Explanation:
To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.
The potential energy 
of the compressed spring is given by
,
where 
is the length of compression and 
is the spring constant.
And the kinetic energy of the ball is
When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,
solving for 
we get:
And since momentum of the ball is 
,
Putting in numbers we get:
Answer:
All fraction of kinectic energy is lost to barrel of a spring gun of mass 1.8 kg
Explanation:
A ball of mass 0.50 kg is fired with velocity 160 m/s ...
The kinetic energy is given by 1/2mv²
Kinectic energy of the ball = 1/2 *0.5*160²
Kinectic energy = 1/4 *25600
Kinectic energy = 6400 joules.
If no energy is lost to fiction, and the ball sticks to a barrel of a spring gun of mass 1.8 kg with initial velocity zero, all kinetic energy is lost to the barrel of a spring gun of mass 1.8 kg.
In vacuum, going at 2.99×10^8 m/s.
Answer:
C. 3.00 s
Explanation:
Given:
Δy = 1.80 m − 46.0 m = -44.2 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 3.00 s
Initial angular velocity = 0 (starting from rest)
Final angular velocity = 30 rad/s
Distance traveled = (20 rev)*(2π rad/rev) = 40π rad
Let the angular acceleration be α rad/s².
Then
(30 rad/s)² = (0 rad/s)² + 2*(α rad/s²)*(40π rad)
α = 3.58 rad/s²
Answer: 3.58 rad/s² (nearest hundredth)
